$\displaystyle f(x) = \dfrac{\sqrt{x^2 + 2x - 8}}{\sqrt{x^2 - x - 12}} = \dfrac{\sqrt{(x+4)(x-2)}}{\sqrt{(x+3)(x-4)}}$
So, x cannot be equal to -3 or 4, because there, you have an asymptote.
At x = -4 and 2, you have f(x) = 0.
Anything negative is imaginary.
When you take the intervals, you have -4, -3, 2, 4.
And trying in values smaller than -4 gives a solution.
At -4, you have 0.
Between -4 and -3, you don't have any solution.
At -3, you get infinity.
Between -3 and 2, you get solutions.
At 2, you get 0 again.
Between 2 and 4 you get no solutions.
At 4, you get again infinity and beyond that, you get solutions.